A circle with radius $5$ has a sector with a $240^\circ$ central angle. What is the area of the sector? ${25\pi}$ $\color{#9D38BD}{240^\circ}$ ${\dfrac{50}{3}\pi}$ ${5}$
Solution: First, calculate the area of the whole circle. Then the area of the sector is some fraction of the whole circle's area. $A_c = \pi r^2$ $A_c = \pi (5)^2$ $A_c = 25\pi$ The ratio between the sector's central angle $\theta$ and $360^\circ$ is equal to the ratio between the sector's area, $A_s$ , and the whole circle's area, $A_c$ $\dfrac{\theta}{360^\circ} = \dfrac{A_s}{A_c}$ $\dfrac{240^\circ}{360^\circ} = \dfrac{A_s}{25\pi}$ $\dfrac{2}{3} = \dfrac{A_s}{25\pi}$ $\dfrac{2}{3} \times 25\pi = A_s$ $\dfrac{50}{3}\pi = A_s$